Consider the following concentration–time data for the
reaction ...

Question

Consider the following concentration–time data for the reaction of iodide ion and hypochlorite ion 1OCl-2. The products are chloride ion and hypoiodite ion 1OI-2. (b) Determine the rate law, and calculate the value of the rate constant.

Accepted Answer

Well, everyone said in this video, we're given this table here that contains the concentration time data for our reaction. We're being asked to give the rate law and provide the value of the rate constants. So the rate equation is that rate is equal to our key constant, multiplied by the concentration of C L. O minus. Raised to power of M multiplied by the concentration of B. R minus, raised to power of N. Multiplied by the concentration of H 20. Raised to power of P. We're gonna go ahead and determine the initial rates by using this equation here. So just put maybe a star here. So this rate is equal to negative delta, which the change in the concentration of C. L minus over delta T. Which is just also equal to the concentration of C. L. O minus. What is the final value minus? The concentration of R C L. O minus its initial value. All over delta T. All right. So for our experiment one year, which we'll do in red again, where is following the initial rates equation? So the first one here is rate is equal to the change in our cielo minus. We have three 00 Times 10 to the -4 molars -3.250 Times 10 to the -4 Moller. This is all going to be over Delta T. And that's just 30 seconds. So, we see here for experiment one, we're dealing with this here. So we're going from 30 or zero seconds to 30 seconds and then we're going from this final concentration to this initial concentration. So that's why we have these values in here and we're just basing it off of this equation here. So the concentration of our final minus concentration of initial over the change over time. Alright, now moving on to experiments to here, the rate is equal to negative 6.0 times 10 to the negative four moller minus 7.0 times 10 to the negative four Moeller. Again, this all over the change in time, which I believe all of them should be 30 seconds. You'll see here that it keeps going from 0 to 30 0 to 30 and so on. Alright then moving on to our third experiment again, rate is equal to the final temper the final value or the concentration value here, that is three points to five times 10 to the negative four moller, an initial is 3.50 times 10 to the negative four moller. And again, this is all over. A change of time of just 30 seconds. Last month we have going on here is experiment for in purple the right equals to the negative 2.50 times 10 to the negative four moller minus 3.50 times 10 to the negative four molars. And again, this is all over 30 seconds. So now that we have these equations set up, we're gonna go ahead and simply put all these values into the calculator. So starting off with experiment one and we put this in the calculator. I get that my rate is equal to 1.67 times 10 to the negative six units being M per S. So this one here is one of my final answers for my problem. Moving on to experiments to let's actually go ahead and give us more space to write our answers. Alright, so for experiment to want to put this into my calculator, I get that my rate or the initial rate rather is equal to 3.33 times 10 to the negative six units are all the same here in our answers. So this right here is going to be my second answer for this problem. Now for experiment three, want to put that into my calculator, I get 8.33 times 10 to the negative seven molar per second. And our last and final answer for experiment four is that the initial rates is equal to 3.33 times 10 to the negative six molar per second. All right, so this is going to be my final answer for this portion. Now let's go ahead and scroll down that we're being asked to provide the values of our rate constants. Alright, so for experiments one and two, the concentration of C. L. O minus doubles. And you can see that the initial rate will also double. Therefore we know that the exponents of M is equal to one. Now for experiment one and three to compare those, the concentration of B. R minus is hat and the initial rates is also halved. Therefore the our exponent of N is equal to one. And lastly for comparing experiments one and 4, The concentration of H. 2 0 doubles and the initial rate doubles. So this means that our p value is equal to one. So putting this into our rate formula. Again that is rate is equal to R. K. Constant, multiplied by the concentration of cielo minus, multiplied by R. B. R minus, multiplied by the concentration of H. 20. So we put in all the values of our exponents. So 11 and one which we all really need to write out. We kind of assume that already. So from experiment one, the K value is equal to on top is gonna be 1.67 times 10 to the negative six molar per second. And my denominator we have 3.50 times 10 to the negative four moller times 4.0 times 10 to the negative four molars times one molar. So once you put that into the calculator you get the K. is equal to 11.9 M. to the -2 times s two negative one. So moller to negative two and seconds to negative one. So this is our final answer for this problem. Thank you so much for watching